MCQ
$\Delta H_f^o$ of water is $-\,285.5\,kJ\,mol^{-1}$ . if enthalpy of neutralisation of monoacidic strong base is $-57.3\,kJ\,mol^{-1},$ of $OH^-$ ion will be.....$kJ\,mol^{-1}$
- A$-228.2$
- B$228.5$
- C$114.5$
- D$-114.5$
✓
Answer
$H^+(a q) OH ^{-}(a q)+OH^{-}(a q) \rightarrow H _2o(l)$
$\text { so } \Delta H^{\circ}=-57.3=-285.5-\Delta_f^{\circ}(H+a q)-\Delta H_f^{\circ}\left(OH^{-}\right)$
so, $\Delta H^o_{f} (OH^- aq)=-228.2\, kJ / mol$
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